Y is a subspace of a X which is a topology

• Sep 23rd 2011, 07:51 AM
dwsmith
Y is a subspace of a X which is a topology
Let Y be a subspace of X. If A is closed in Y and Y is closed in X, then A is closed in X.

First are there 2 cases to consider? (1) A is a proper subset of Y (2) \$\displaystyle A=Y\cap U\$ or does this not matter?

Anyways, all I have is this:

Since A is closed in Y, \$\displaystyle Y-A\$ is open in Y. Likewise, \$\displaystyle X-Y\$ is open in X.

I am not sure what to do next.
• Sep 23rd 2011, 10:23 AM
Drexel28
Re: Y is a subspace of a X which is a topology
Quote:

Originally Posted by dwsmith
Let Y be a subspace of X. If A is closed in Y and Y is closed in X, then A is closed in X.

First are there 2 cases to consider? (1) A is a proper subset of Y (2) \$\displaystyle A=Y\cap U\$ or does this not matter?

Anyways, all I have is this:

Since A is closed in Y, \$\displaystyle Y-A\$ is open in Y. Likewise, \$\displaystyle X-Y\$ is open in X.

I am not sure what to do next.

By definition that \$\displaystyle A\$ is closed in \$\displaystyle Y\$ you have that \$\displaystyle A=Y\cap C\$ for some closed subset \$\displaystyle C\$ of \$\displaystyle X\$....so.
• Sep 23rd 2011, 10:35 AM
dwsmith
Re: Y is a subspace of a X which is a topology
Quote:

Originally Posted by Drexel28
By definition that \$\displaystyle A\$ is closed in \$\displaystyle Y\$ you have that \$\displaystyle A=Y\cap C\$ for some closed subset \$\displaystyle C\$ of \$\displaystyle X\$....so.

I don't know. All I can think of is to do \$\displaystyle Y-(Y\cap A)\$ then but I don't see how that will help.
• Sep 23rd 2011, 07:51 PM
dwsmith
Re: Y is a subspace of a X which is a topology
Ok, I think I have it.

\$\displaystyle A=Y\cap U\$ where U is some closed set in X. Now, A is the intersection of two closed sets in X so A is closed in X.
• Sep 23rd 2011, 07:52 PM
Drexel28
Re: Y is a subspace of a X which is a topology
Quote:

Originally Posted by dwsmith
Ok, I think I have it.

\$\displaystyle A=Y\cap U\$ where U is some closed set in X. Now, A is the intersection of two closed sets in X so A is closed in X.

Correct.